Question: Find the zeros of the function. Enter the solutions from least to greatest. $f(x) = (x - 10)^2 - 49$ $\text{lesser }x = $
$\begin{aligned} (x - 10)^2 - 49&= 0 \\\\ (x-10)^2&=49 \\\\ \sqrt{(x-10)^2}&=\sqrt{49} \end{aligned}$ $\begin{aligned} x-10&=\pm7 \\\\ x&=\pm7+10 \\ \phantom{(x - 10)^2 - 49}& \\ x=3&\text{ or }x=17 \end{aligned}$ In conclusion, $\begin{aligned} \text{lesser }x &= 3 \\\\ \text{greater } x &= 17 \end{aligned}$